**Given:**

The 17^{th} term of an A.P. is 5 more than twice its 8^{th} term and the 11^{th} term of the A.P. is 43.

**To find:**

the n^{th} term.

**Solution:**

Given, a_{17} = 5 + 2(a_{8}) ......(i)

And a_{11} = 43

As a_{n} = a + (n -1) d

For n = 11,

a + 10d = 43a = 43 - 10d ......(ii)

For n = 8, a_{8} = a + 7d

For n =17,

a_{17} = a + 16d

**Put the value of a from (ii)**

⇒ a_{17}= 43 – 10d + 16d

⇒ a_{17} = 43 + 6d

**Putting the value of a**_{8} and a_{17} in (i), we get

⇒ 43 + 6d = 5 + 2(a + 7d)

⇒ 43 + 6d = 5 + 2a + 14d

⇒ 43 – 5 = 2a + 14d – 6d

⇒ 38 = 2a + 8d

⇒ 38 = 2(43 – 10d) + 8d [from (ii)]

⇒ 38 = 86 – 20d + 8d

⇒ 38 = 86 – 12d

⇒ 12d = 86 – 38

⇒ 12d = 48

⇒ d = 4

**From (ii),**

a = 43 – 10d

⇒ 43 – (10 × 4) = 43 – 40 = 3

**We know, n**^{th} term of A.P., a_{n} = a + (n -1) d

a_{n }= 3 + (n – 1) 4

a_{n} = 3 + 4n – 4 a_{n} = 4n – 1

**Hence, n**^{th} term is 4n – 1.